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3 years ago

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What is the shape of the cross-section formed when a plane containing line AC and line EH intersects this cube? What is the area

of the cross-section? In two or more complete sentences, explain how you were able to determine the shape and area of the cross-section. Be sure to include the properties of quadrilaterals and the mathematical computations necessary to support your answers

1 answer:

velikii [3]3 years ago

4 0

Answer:

Cross section is a rectangle.

Area of cross section = 16 sqrt(2) = 22.63 sq. units (to 2 decimals)

Step-by-step explanation:

Given a cube.

top face ABCD is parallel and congruent to bottom face EFGH ........(1)

justified by the properties of cubes

Sides AE and CH are perpendicular to faces ABCD and EFGH ..........(2)

justified by the properties of cubes

Diagonals AC and EH are congruent ......................(3)

justified by (1), congruent top and bottom faces

Consider cross-section ACHE

AC is congruent and parallel to EH (1) & (3)

EA & HC are perpendicular to AC (2)

Therefore the quadrilateral ACHE is a rectangle.

Length of diagonal AC = sqrt(4^2+4^2) = 4 sqrt(2) ..........pythagoras theorem

AE = CH = DG = 4 properties of cube, all sides equal

Area of ACHE = 4* 4sqrt(2) = 16 sqrt(2) = 22.63 sq. units

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Could someone help me rnnn?

GuDViN [60]

Answer:

vertex = (0, -4)

equation of the parabola: $y=3x^2-4$

Step-by-step explanation:

Given:

y-intercept of parabola: -4
parabola passes through points: (-2, 8) and (1, -1)

Vertex form of a parabola: $y=a(x-h)^2+k$

(where (h, k) is the vertex and $a$ is some constant)

Substitute point (0, -4) into the equation:

$\begin{aligned}\textsf{At}\:(0,-4) \implies a(0-h)^2+k &=-4\\ah^2+k &=-4\end{aligned}$

Substitute point (-2, 8) and $ah^2+k=-4$ into the equation:

$\begin{aligned}\textsf{At}\:(-2,8) \implies a(-2-h)^2+k &=8\\a(4+4h+h^2)+k &=8\\4a+4ah+ah^2+k &=8\\\implies 4a+4ah-4&=8\\4a(1+h)&=12\\a(1+h)&=3\end{aligned}$

Substitute point (1, -1) and $ah^2+k=-4$ into the equation:

$\begin{aligned}\textsf{At}\:(1.-1) \implies a(1-h)^2+k &=-1\\a(1-2h+h^2)+k &=-1\\a-2ah+ah^2+k &=-1\\\implies a-2ah-4&=-1\\a(1-2h)&=3\end{aligned}$

Equate to find h:

$\begin{aligned}\implies a(1+h) &=a(1-2h)\\1+h &=1-2h\\3h &=0\\h &=0\end{aligned}$

Substitute found value of h into one of the equations to find a:

$\begin{aligned}\implies a(1+0) &=3\\a &=3\end{aligned}$

Substitute found values of h and a to find k:

$\begin{aligned}\implies ah^2+k&=-4\\(3)(0)^2+k &=-4\\k &=-4\end{aligned}$

Therefore, the equation of the parabola in vertex form is:

$\implies y=3(x-0)^2-4=3x^2-4$

So the vertex of the parabola is (0, -4)

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